Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import javax.swing.*;
import java.io.*;
import java.sql.Struct;
import java.text.DecimalFormat;
import java.util.*;

public class Main {


    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        double[] board = new double[n];
        for(int i = 0; i < n; i++) {
            board[i] = sc.nextDouble();
        }
        double[][] probabilities = new double[n][n + 1];
        probabilities[0][0] = board[0];
        probabilities[0][1] = 1 - board[0];

        for(int i = 1 ; i < n; i++) {
            for (int j = 0; j <= n; j++) {
                double res = 0;
                if (i > 0) {
                    res += probabilities[i - 1][j] * board[i];
                }
                if (j > 0 && i > 0) {
                    res += probabilities[i - 1][j - 1] * (1 - board[i]);
                }
                probabilities[i][j] = res;
            }
        }

        double res = 0;
        for (int j = 0; j <= n / 2 ; j++) {
            res += probabilities[n - 1][j];
        }

        System.out.println(String.format("%.9f", res));
    }
}